5- Sept- 2014 Elephant Example

Elephant Example:

Objective:

  Show the difference to solving a question with a non- constant acceleration numerically vs. using calculus to solve the problem.

Question:

 A 5000-kg elephant on frictionless roller skates is going 25 m/s when it gets to the bottom of a 

hill and arrives on level ground. At that point a rocket mounted on the elephant’s back generates
a constant 8000 N thrust opposite the elephant’s direction of motion.
The mass of the rocket changes with time (due to burning the fuel at a rate of 20 kg/s) so that the
m(t) = 1500 kg – 20 kg/s·t.

Processes:
 Here we will compare the two ways of solving this question starting with using calculus to find our distance of the elephant when it stops. It is apparent that mass isn't constant and there is a constant unequal force pushing horizontally in the x- direction, therefore, there will be a non- constant accelerating elephant.

Fnet = a(t)

Mtotal        

=          (-8000N)           

(5000 kg + (1500 kg + 20 Kg/s (t)) 

Analytical Process:

 From our equation set as our a(t), we see that we must find the the change in velocity (Δv) in order to find v(t) to eventually find the time when v(t)= 0m/s. To do this, we add 
our initial 25m/s to our integral from 0 to t of the acceleration a(t). After finding our v(t), 
we tried to adding xo to the integral of our velocity from 0 to t, similarly to getting v(t) only
this time for x(t). Work shown in the figure below (fig 1.)

Figure 1. Shows my revised work for photo clarity purposes. I labeled that I started the steps as

 was described by the paragraph prior respectively. We get that our distance for the elephant when

 v= 0 m/s is at x(19.608)= 248.7m at t = 19.608 s.

Numerically:

 To show another way to solve this problem, we were instructed to use our Excel program and to label our columns as shown below, as seen in (fig 2.)  The columns are labeled as time (t), instantaneous acceleration (a), average of the two accelerations of the time interval (a_avg), change in velocity between two velocities in a time interval (Δv), instantaneous velocity (v), change in distance during a 

time interval (Δx), and total distance (x) respectively. We first decided how far apart our time intervals

would be, in this case it was decided to have a Δt of .1s. 

Figure 2. The first few rows of our Excel spreadsheet described by the previous paragraph. We arranged the columns in an order convenient to the process of which variables are being calculated first all the way down to the variable we mean to solve. Similar to the analytical process.

Using the equation we found on acceleration equaling the net force over mass at a given time, we had our acceleration column calculate the instantaneous acceleration a(t) for each value of t until a certain time (we chose t from 0s to 30s). From there, we found the average of the acceleration (a_avg):

a_avg= (at + at+1)

          2

Next we numerically find Δv :

Δv= (a_avg)* (Δt)

Now that we found our values of Δv we then find v(t) for each value of t by:

v(t)= vn +(vn - 1)  

Using our v(t) values, we then try to find the values of x(t) by using v_avg between two velocities
 multiplied by Δt: 
  
                                                        x(t)= (vn + vn - 1)*(tn - tn - 1 ) + xo
                                                                      2

And finally, for analysis, we used x(t) values to find the Δx by:

                                                             Δx= xn + 1 - xn

As a result for this numerical set -up, we find that v(t) becomes zero between 19.6s < t < 19.7s
because the instantaneous velocities in 19.6s is negative and at 19.7s is positive. The distance at
which the elephant is in those times are very close to what our solution is in the analytical process,
as seen in (fig 3.)

Figure 3. Shows our numerical calculations from 19.5s to 19.7s highlighted since that is the area in which v is found to have been zero between times 19.6s and 19.7s.

We will use the average of distance between the two times to get a rough value in which the elephant stops:

x_avg= x(19.6) + x(19.7) = 248.6953m

2

When we do the same process, but in increments of .05s we see in (fig 4.) that the  our x_avg is slightly different than if we had a tenth of a second intervals, in fact it is closer to our analytical answer than the former by only a slight margin due to smaller time intervals.

x_avg= 248.6976m

Figure 4. The same numerical set- up but with increments Δt= .05s, resulting to a more detailed

 spreadsheet on elephant's scenario variable- wise.

However, if we changes the intervals to Δt = 1s (as shown in fig 5.) we have a less accurate spreadsheet than if we had 

Δt= .1s or .05s. Now the average betweens t= 19s to t= 20s is:

                                                             x_ avg= 248.5037m

Figure 5. Our spreadsheet with the same set- up but with intervals of Δt= 1s which has shown to be

less detailed as the former two figures.

Conclusion: 

 Now if we compare all of our results for finding what distance the elephant stops :

Analytical:                           x=  248.7m

Numerical:  Δt= .1s       x_ avg= 248.6953m     -.0047 m from analytical

                 Δt= .05s      x_ avg= 248.6976m     -.0024 m from analytical

                Δt= 1s          x_ avg= 248.5037m     -.1963 m from analytical 

We choose our time intervals for our numerical process to be smaller and smaller until the difference between the two resulting values in question is negligible. In this case even choosing the time in tenths of a second would have sufficed seeing how its difference between the two times in question is negligible and is only approximately 5 thousandths of a meter difference from our analytical answer. Also in return, doing these sort of problems help be used as references for when we solve the problem analytically. They both use similar processes of using acceleration first then velocity then find position and in the end (assuming your numerical calculations are correct) it can show if you resulted to the wrong analytical answer.